91Ó°ÊÓ

In chemical reactions, the equilibrium constant \( K_c \) is vital for understanding how far a reaction will proceed before achieving a balance where no net change occurs. In our exercise, we’re dealing with a chemical reaction between carbon monoxide (CO) and chlorine (Cl\(_2\)) to form phosgene (COCl\(_2\)). The reaction forms a dynamic state of balance, measurable by \( K_c \), which relates the concentrations of reactants and products at equilibrium. Here, the balanced equation is:\[ \text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g) \]The equilibrium constant expression for this reaction is given by:\[ K_c = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} \]This means \( K_c \) is the ratio of the concentration of the product to the concentrations of the reactants, raising them to the power of their stoichiometric coefficients (which are all 1 in this case). In our problem, \( K_c \) is specified as 2.5, indicating that at equilibrium, the product's concentration in the container slightly outweighs the concentrations of the reactants.

Calculating concentrations at equilibrium involves some basic mathematical operations that arise naturally from the definition of \( K_c \). Initially, we determine the concentrations of each substance in the reaction:

• For CO: \( \frac{1.00 \text{ mol}}{3.00 \text{ L}} = 0.333 \text{ mol/L} \)
• For Cl\(_2\): \( \frac{2.00 \text{ mol}}{3.00 \text{ L}} = 0.667 \text{ mol/L} \)
• COCl\(_2\) initially is 0 mol/L since the reaction hasn’t progressed.

When solving for equilibrium conditions, assume a certain change \( x \) in concentrations occurs. For this reaction, the equilibrium concentrations are:

• \([\text{CO}] = 0.333 - x\)
• \([\text{Cl}_2] = 0.667 - x\)
• \([\text{COCl}_2] = x\)

This step is crucial, as by substituting these into the \( K_c \) equation, we can solve for \( x \), which tells us exactly how much COCl\(_2\) has formed when the reaction reaches equilibrium.

Understanding the mole concept helps us link the amounts of substances with their concentrations in a given volume. A mole represents a specific quantity of particles, which in chemistry, denotes the number of atoms, molecules, or ions in a substance. In this exercise, we start with 1.00 mol of CO and 2.00 mol of Cl\(_2\), enclosed in a container with a 3.00 L volume. By using the formula:\[ \text{Concentration} = \frac{\text{moles}}{\text{volume}} \]we translate these amounts into molar concentrations:

• 1.00 mol of CO becomes \( 0.333 \text{ mol/L} \)
• 2.00 mol of Cl\(_2\) becomes \( 0.667 \text{ mol/L} \)

The initial concentration of COCl\(_2\) is zero, as no reaction has occurred yet. This translation sets the stage for us to consider the changes in concentration that happen during the reaction and to use the equilibrium constant \( K_c \) to predict the equilibrium concentrations. Hence, by using the mole concept, we perform essential calculations needed to solve equilibrium problems.